Question: Divide the following complex numbers. $ \dfrac{3+21i}{3-3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+3i}$ $ \dfrac{3+21i}{3-3i} = \dfrac{3+21i}{3-3i} \cdot \dfrac{{3+3i}}{{3+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(3+21i) \cdot (3+3i)} {(3-3i) \cdot (3+3i)} = \dfrac{(3+21i) \cdot (3+3i)} {3^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(3+21i) \cdot (3+3i)} {(3)^2 - (-3i)^2} = $ $ \dfrac{(3+21i) \cdot (3+3i)} {9 + 9} = $ $ \dfrac{(3+21i) \cdot (3+3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({3+21i}) \cdot ({3+3i})} {18} = $ $ \dfrac{{3} \cdot {3} + {21} \cdot {3 i} + {3} \cdot {3 i} + {21} \cdot {3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{9 + 63i + 9i + 63 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{9 + 63i + 9i - 63} {18} = \dfrac{-54 + 72i} {18} = -3+4i $